Suppose we wanted to evaluate the double integral $S = \iint_D -x + 3y \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = -v^3 + u^2 - v \\ \\ y &= X_2(u, v) = 2u \end{aligned}$ What is $S$ under the change of variables? If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
Explanation: If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} 2u & -3v^2 - 1 \\ \\ 2 & 0 \end{pmatrix} \right| \\ \\ &= \left| -6v^2 - 2 \right| \\ \\ &= 6v^2 + 2 \end{aligned}$ Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= -X_1(u, v) + 3X_2(u, v) \\ \\ &= -(-v^3 + u^2 - v) + 3(2u) \\ \\ &= v^3 - u^2 + v + 6u \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R (6v^2 + 2)(v^3 - u^2 + v + 6u) \, du \, dv$